I have a degree in math and I actually enjoy these types of problems. Keeping sharp at mental arithmetic is really fun and it prompts the brain to come up with shortcuts that can lead to more interesting math. Like, is 51 prime? (No, 5+1 = 6). It's not really rote memorization but getting better at mental arithmetic, which is actually meditative in a way.
For any integer base B greater than 2, the multiples of the number B-1, when represented in base B will always have the sum of the digits be a multiple of B-1. If B-1 is a square number (4,9,16,...) then the square root of B-1 will also have this property.
> If B-1 is a square number (4,9,16,...) then the square root of B-1 will also have this property.
In fact, if d is any divisor of B-1 (including a trivial divisor), then d will have this property. (The author of the proof on the page you linked to is aware of this, but the author of the original conjecture isn't. For example, the reply notes that the digital root test for divisibility in hexadecimal works for 3 or 5, which are divisors of 16-1=15, regardless of the fact that 15 isn't a square. It also works for 1 or 15, which are also divisors of 15.)
